For any space $X$, the first Steenrod square cohomology operation
$$Sq^1\colon H^\ast(X;\mathbb{Z}_2)\to H^{\ast +1}(X;\mathbb{Z}_2)$$
is a derivation, meaning that $Sq^1\circ Sq^1 = 0$ and $Sq^1(a\cup b) = Sq^1(a)\cup b + a\cup Sq^1(b)$ (there are no signs since we are working in characteristic two).

Hence we may form the $Sq^1$-cohomology of the space,
$$H\left(H^\ast(X;\mathbb{Z}_2),Sq^1\right)$$
which will be a graded algebra over $\mathbb{Z}_2$.

I am looking for references on this object. From McCleary's "User's guide to spectral sequences", I know that this is related to the Bockstein spectral sequence. More specifically, I would like to know:

What is the precise relationship between the $Sq^1$-cohomology of a space $X$ and $2$-torsion of higher order in $H^\ast(X;\mathbb{Z})$?

Is there a reference with specific calculations of the $Sq^1$-cohomology of the Eilenberg-Mac Lane spaces $K(\mathbb{Z}_2,n)$?

Are there any canonical references I should know about (besides McCleary and Mosher-Tangora)?

i think this is also related to Margolis homology. you might want to look at that. The issues there are more related to classification of modules over subalgebras of the steenrod algebra though.
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Sean TilsonAug 28 '11 at 21:31

4 Answers
4

I think the easiest way to understand the Bockstein spectral sequence is through the exact couple coming from the long exact sequence of cohomology associated to $0\to\mathbb Z\to\mathbb Z\to \mathbb Z/2\to0$. This shows first that indeed the first differential is $Sq^1$ and tells you that the next page is the direct sum of the cokernel and kernel (shifted one step) of multiplication by $2$ on $2H^\ast(X,\mathbb Z)$. Hence it is like what you would get from applying the universal coefficient formula to $2H^\ast(X,\mathbb Z)$ (instead of $H^\ast(X,\mathbb Z)$). When each cohomology group $H^\ast(X,\mathbb Z)$ is finitely generated this means concretely that you "keep" each $\mathbb Z$-factor (as well as odd torsion) and downgrade each $\mathbb Z/2^n$ to $\mathbb Z/2^{n-1}$.

In particular the difference between the dimension of $H^n(X,\mathbb Z/2)$ and that of the $Sq^1$-cohomology is equal to the number of $\mathbb Z/2$-factors in $H^n(X,\mathbb Z)$ and $H^{n+1}(X,\mathbb Z)$.

I found a reference to Q2. In Madsen, Milgram: The classifying spaces for surgery and cobordism of manifolds, Ann of Math Studies 92 where they refer to Browder: Torsion in H-spaces, Ann of Math 74 for the Bockstein s.s. of $K(\mathbb Z_{(2)},n)$ and $K(\mathbb Z/2,n)$. The Madsen-Milgram book also contains other examples of computations with the Bss.

Hi Torsten. This was helpful, thanks. There are some words missing from the last sentence of your first paragraph, perhaps "finitely generated"?
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Mark GrantAug 29 '11 at 8:13

Thanks again. The Browder paper looks like the reference I was seeking.
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Mark GrantAug 29 '11 at 15:34

This really does answer all 3 of my questions! For anyone interested, the Bockstein SS for $K(Z_2,n)$ is given a fairly explicit description in Theorem 5.5 of the Browder paper.
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Mark GrantAug 30 '11 at 9:24

I remember that when I wrote my thesis I was unable to find references for some quite basic facts about this that everyone knew. It is quite possible that there were references that I did not succeed in finding (there was no Google then!) or that someone has written a good exposition in the intervening time. For what it's worth, my thesis is at http://neil-strickland.staff.shef.ac.uk/research/thesis.pdf and the relevant material is in Section 5.1.

Several people have addressed question 1 (Torsten Ekedahl and Neil Strickland). Question 2 is interesting, but I don't have a good answer for it. For question 3, as Sean Tilson points out, this is a special case of "Margolis homology", a.k.a. $P^s_t$-homology. Try

Bill Singer and I have looked at the "dual" question... Given a right $\mathcal{A}$-module $M$ (such as the homology of a space), $Sq^1$ acts (on the right) as a differential. For any $s \geq 1$ and $M = \tilde{H}_*((\mathbb{R}P^{\infty})^{\wedge s}, \mathbb{Z}/2)$, $Sq^1$-homology is trivial, indicating that the kernel of $Sq^1$ is the same as the image of $Sq^1$ there. There are interesting things to say even for higher squares, though they won't be differentials in general.

The following references may be useful (both available on arXiv):

Ault, Singer. On the Homology of Elementary Abelian Groups as Modules over the Steenrod Algebra.

Ault. Relations among the kernels and images of Steenrod squares acting on right $\mathcal{A}$-modules.